Continuous-time Fourier transform computation (in terms of frequency f in hertz)

Compute the Continuous-time Fourier transform of the two following functions:

$x(t)= \text{rect}(t) = \left\{ \begin{array}{ll} 1, & \text{ if } |t|<\frac{1}{2}\\ 0, & \text{ else} \end{array} \right.$

$y(t)= \frac{ \sin ( \pi t )}{\pi t}$

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Answer 1

Fourier Transform of rect(t):

$X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dx =\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dx =\frac{e^{-j2\pi ft}}{-j2\pi f}$ from t=-1/2 to t=1/2

$=\frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} =\frac{sin(\pi f)}{\pi f}$

Instructor's comments: Technically, you should look at the case f=0 separately, because your solution involves a division by f. -pm

Fourier Transform of $\frac{sin(\pi t)}{\pi t}$ :

Guess: $$ X(f)=rect(t) $$

Proof:

$x(t)=\int_{-\infty}^{\infty} X(f)e^{j2\pi ft} df =\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{j2\pi ft} df =\frac{e^{j2\pi ft}}{j2\pi t}$ from f=-1/2 to f=1/2

$=\frac{e^{j\pi t}-e^{j\pi t}}{j2\pi t} =\frac{sin(\pi t)}{\pi t}$

Instructor's comments: Guessing the answer and proving it using the inverse Fourier transform is a good trick. One could also obtain this Fourier transform using the duality property and your previous answer. -pm

Answer 2

$X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt =\int_{-\infty}^{\infty} rect(t)e^{-j2\pi ft} dt =\int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt$

$= -\frac{e^{-j2\pi ft}}{-j2\pi f}$ integrating from -0.5 to +0.5. $= \frac{e^{-j\pi f} - e^{j\pi f}}{-j2\pi f} = \frac{sin(\pi f)}{\pi f}$

For y(t), we know that

$y(t) = \int_{-\infty}^{\infty} Y(f)e^{j2\pi ft} df$

$y(t)= \frac{ \sin ( \pi t )}{\pi t} = \frac{e^{-j\pi t} - e^{j\pi t}}{\pi t}$

For the above equation to be true,

$Y(f) = \frac{\delta(f - \frac{1}{2})}{\pi t} - \frac{\delta(f + \frac{1}{2})}{\pi t}$

Answer 3

$X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt = \int_{-1/2}^{1/2} e^{-j2\pi ft} dt = \frac{e^-j2\pi ft}{j2 \pi f},where x from-1/2 to 1/2 =sinc(f)$ $$ use duality, Y(f)=rect(x) $$

Answer 4

Using Euler's equation, we know

$y(t)= \frac{ \sin ( \pi t )}{\pi t} = \frac{e^{j\pi t} - e^{-j\pi t}}{j2\pi t}$

Do Fourier transform of rect(t),

$\begin{align}\mathcal{F}[\text{rect}(t)] = \int_{-\infty}^{\infty} \text{rect}(t)e^{-j2\pi ft} dt \\ = \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt \\ = \frac{e^{-j2\pi ft}}{-j2\pi f}\vert \end{align}$ evaluate from -1/2 to 1/2,

$\begin{align} Y(f)=\frac{e^{j\pi f}-e^{-j\pi f}}{j2\pi f} =\frac{sin(\pi f)}{\pi f} \end{align}$

use duality, $$ Y(f)=rect(x) $$

Answer 5

Since the function is periodic we can integrate over one period:

F[rect(t)] = \frac{1}{1} \int_{-\frac{1}{2}}^{\frac{1}{2}}\ (1) e^{j2\pi (1)t} df

= \frac{1}{-j2\pi} e^{-j2\pi t} \Bigg|_{-\frac{1}{2}}^{\frac{1}{2}} = \frac{-1}{j2\pi k } (e^{-j2\pi f\frac{1}{2}} - e^{j2\pi f\frac{1}{2}})

= \frac{sin(\pi f)}{\pi f}

To find the FT of the sinc function, simply use duality from the first solution:

$$ F[sinc(t)] = rect(-f) $$

and since $$ rect $$ is even,

$rect(-f)= \text{rect}(f) = \left\{ \begin{array}{ll} 1, & \text{ if } |f|<\frac{1}{2}\\ 0, & \text{ else} \end{array} \right.$

Answer 6

$\begin{align} X(f)&=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dx \\ &=\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dx \\ &=\frac{e^{-j2\pi ft}}{-j2\pi f} \big| _{-1/2}^{1/2} \\ &=\frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} \\ &=\frac{sin(\pi f)}{\pi f} \end{align}$

duality:

$f(t) \Leftrightarrow F(-\omega)$

$F(t) \Leftrightarrow 2\pi f(-\omega)$

so

$rect(t) \Leftrightarrow sinc(\omega/2)$

$2 \pi sinc(t) \Leftrightarrow 2 \pi rect(-\omega/2\pi) = rect(f)$

Answer 7

Start with x(t) = rect(t)

$X(f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt$

$= \int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt$

$= \frac{e^{-j2\pi ft}}{-j2\pi f} \Bigg| _{-1/2}^{1/2}$

$= \frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f}$

$= \frac{ \sin (\pi f)}{\pi f}$

Now the FT of the sinc function is much easier. Use Duality to show:

$$ F{(sinc(t))} = rect(-f) = rect(f) $$

Answer 8